Problem 62
Question
Solve each equation, and check the solution. \(\frac{8 x}{3}-\frac{x}{2}=-13\)
Step-by-Step Solution
Verified Answer
The solution is x = -6.
1Step 1: Find a common denominator
To combine the fractions \(\frac{8x}{3}\) and \(\frac{x}{2}\), find a common denominator. The smallest common multiple of 3 and 2 is 6.
2Step 2: Express each term with the common denominator
Rewrite each term to have a denominator of 6: \(\frac{8x}{3} = \frac{16x}{6}\) and \(\frac{x}{2} = \frac{3x}{6}\).
3Step 3: Combine the fractions
Rewrite the equation as \( \frac{16x}{6} - \frac{3x}{6} = -13 \) and combine the terms on the left side: \( \frac{13x}{6} = -13 \).
4Step 4: Solve for x
To isolate x, multiply both sides by 6: \( 13x = -78 \). Then divide both sides by 13: \( x = -6 \).
5Step 5: Check the solution
Substitute \( x = -6 \) back into the original equation to verify: \(\frac{8(-6)}{3} - \frac{(-6)}{2} = -13 \).Simplify: \(-16 + 3 = -13\).Since both sides are equal, \( x = -6 \) is correct.
Key Concepts
common denominatorsfractions in algebrachecking solutions in equations
common denominators
When solving equations with fractions, finding a common denominator is crucial. It allows us to combine fractions easily.
To find the common denominator, identify the least common multiple (LCM) of the denominators. In our example, the denominators are 3 and 2.
The LCM of 3 and 2 is 6. This means we'd rewrite each fraction with 6 as the denominator.
So, \(\frac{8x}{3}= \frac{16x}{6} \) and \(\frac{x}{2} = \frac{3x}{6} \). From here, combining the fractions becomes straightforward.
To find the common denominator, identify the least common multiple (LCM) of the denominators. In our example, the denominators are 3 and 2.
The LCM of 3 and 2 is 6. This means we'd rewrite each fraction with 6 as the denominator.
So, \(\frac{8x}{3}= \frac{16x}{6} \) and \(\frac{x}{2} = \frac{3x}{6} \). From here, combining the fractions becomes straightforward.
fractions in algebra
Handling fractions in algebra can seem tricky, but with practice, it gets easier. Key steps include finding common denominators and combining like terms.
In the given equation, we started with \( \frac{8x}{3} - \frac{x}{2} = -13 \).After adjusting to the common denominator, the fractions were rewritten: \( \frac{16x}{6} - \frac{3x}{6} \).
Now, they shared the same denominator, so combining them was easy. The result was \( \frac{13x}{6} = -13 \).
From here, converting to a simpler form, we isolated \( x \) by multiplying each side by 6 and then solving for \( x \).
In the given equation, we started with \( \frac{8x}{3} - \frac{x}{2} = -13 \).After adjusting to the common denominator, the fractions were rewritten: \( \frac{16x}{6} - \frac{3x}{6} \).
Now, they shared the same denominator, so combining them was easy. The result was \( \frac{13x}{6} = -13 \).
From here, converting to a simpler form, we isolated \( x \) by multiplying each side by 6 and then solving for \( x \).
checking solutions in equations
After finding a solution for an equation, it's essential to check if it's correct. This verification ensures no errors were made during calculations.
For our equation, we found \( x = -6 \). We substituted \( x = -6 \) back into the original equation to check: \( \frac{8(-6)}{3} - \frac{-6}{2} = -13 \).
Simplifying, we had \( -16 + 3 = -13 \).
Both sides matched, confirming that our solution \( x = -6 \) was indeed correct. Always remember this step, as it helps catch possible mistakes and reinforces understanding.
For our equation, we found \( x = -6 \). We substituted \( x = -6 \) back into the original equation to check: \( \frac{8(-6)}{3} - \frac{-6}{2} = -13 \).
Simplifying, we had \( -16 + 3 = -13 \).
Both sides matched, confirming that our solution \( x = -6 \) was indeed correct. Always remember this step, as it helps catch possible mistakes and reinforces understanding.
Other exercises in this chapter
Problem 62
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Solve each problem. Randall wants to mix 50 lb of nuts worth \(\$ 2\) per \(\mathrm{lb}\) with some nuts worth \(\$ 6\) per \(\mathrm{lb}\) to make a mixture wo
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